# Uniform Converges of Continuous Increasing Functions

1. Jan 11, 2009

### e(ho0n3

The problem statement, all variables and given/known data
Let f, f1, f2, f3, ... be continuous real-valued functions on the compact metric space E, with f = lim fn. Prove that if fi ≤ fj whenever i ≤ j, then f1, f2, ... converges uniformly.

The attempt at a solution
I was trying to reverse engineer the proof, but I'm stuck trying to figure out how the hypothesis, viz. fi ≤ fj whenever i ≤ j, comes into play. Any tips?

2. Jan 11, 2009

### tim_lou

I assume you mean f = lim fn in the point-wise limit, and you want to show that
lim sup|f-f_n| = 0

I thought about it a little bit more, and I think a better approach would be to look at
gn(x)=f(x)-fn(x), gn>0.

these maps are continuous, you only need to show that gn converges uniformly to zero.

draw a horizontal line at y=ϵ, your job is to show that eventually for some n, all of gn(x) lies below this line.
This proof becomes simple then as long as you think in terms of contradictions. What if for some x, none of gn(x) lies below this line?

Last edited: Jan 11, 2009
3. Jan 11, 2009

### tim_lou

And also, was f required to be continuous? otherwise, I don't think this proposition works because you can have fn(x)=x^2 on [0,1] that is monotonically nonincreasing (as n increases) and pointwise convergent to a discontinuous function. This convergence obviously cannot be uniform.

4. Jan 11, 2009

### e(ho0n3

Yes, f is continuous (see first post).

Now if for some x, none of gn(x) lie below the line y = ϵ, then gn(x) = f(x) - fn(x) > ϵ for all n. But I think this contradicts the fact that fn converges point-wise to f: we can always find an N such that |f(x) - fn(x)| < ϵ for all n > N. We have not used the fact that f1(x) ≤ f2(x) ≤ ... though.

5. Jan 13, 2009

### tim_lou

The fact that fn are increasing means that:

g(x) are always positive
gn(x) are always decreasing

also, think about it, how can you conclude that |gn(x)| < ϵ when n>n0 after you've chosen a n0? Here, you've implicitly used that inequality.

6. Jan 13, 2009

### e(ho0n3

We have that g1(x) >= g2(x) >= ... and since E is compact, each gn reaches its maximum, say Mn, which means M1 >= M2 >= ... Furthermore, for each n, Mn >= 0 so the sequence M1, M2, ... is a bounded monotonic, and hence convergent, sequence. In effect, for any ϵ > 0, we can find an N such that |gn(x)| <= Mn < ϵ for all n > N, for all x.

That should do it. Thanks a bunch.

7. Jan 14, 2009

### e(ho0n3

I made a mistake in my last post: I'm assuming that M1, M2, ... converges to 0. I've been attempting to prove this without any success. Could it be that M1, M2, ... doesn't necessarily converge to 0? If so, then my proof is ruined...

8. Jan 14, 2009

### Dick

I'm not quite sure what the M's are, but there are three main points in the proof. i) Since the approach to the limit is monotonic, if |f_N(x)-f(x)|<e then |f_n(x)-f(x)|<e for all n>=N. ii) since f_n and f are continuous, if |f_N(x)-f(x)|<e then there is a neighborhood of x, B(x,N,e) where, say, |f_N(y)-f(y)|<2e for y in B(x,N,e). iii) E is compact, so it can be covered by a finite number of these neighborhoods. Can you put all of this together?

9. Jan 14, 2009

### e(ho0n3

Mn = max{gn(x)}. So you're saying that E is the union of B(x_i, N_i, e), for i = 1, 2, ..., n, say. I don't know how to tie that together with i) and ii). Sorry.

10. Jan 14, 2009

### Dick

I don't think Mn=max{gn(x):x in E} is very useful. Pick an e>0. For every x in E there is a N(x) such that |f_n(x)-f(x)|<e for all n>=N. x has a neighborhood where |f_n(x)-f(x)|<2e. E is compact. It's covered by a finite number of these neighborhoods. Isn't that enough for you to find an M such that |f_n(x)-f(x)|<2e for n>=M and all x in E?

11. Jan 14, 2009

### tim_lou

didn't you already prove the result?

you've just shown sup|gn(x)|->0, didn't you?

12. Jan 14, 2009

### e(ho0n3

Let me reiterate as much as I understand of your argument. Fix an x in E and an e > 0. By convergence, there is an N(x,e) such that |f_n(x) - f(x)| < e for all n > N(x,e). Let N > N(x,e). By continuity of f_n and f, there is a d(N,e) such that, where D is the metric on E, if D(y,x) < d(N,e), then |f_N(y) - f_N(x)| < e and |f(y) - f(x)| < e, yielding |f_N(y) - f(y)| < 2e. Let B(x, N, e) be the open ball of radius d(N,e). By compactness, E may be covered by a finite number of the B's, say B(x_i, N_i, e), i=1, 2, ..., n. Thus x belongs to B(x_i, N_i, e) for some i, so |f_{N_i}(x) - f(x)| < 2e. From here, how do I get that |f_n(x) - f(x)| < 2e for all n > N(e), where N(e) depends solely on e.

But I didn't use the fact that E is compact or that the f1 <= f2 <= ..., so I figured the proof must be incorrect.

13. Jan 14, 2009

### Dick

You have a finite number of neighborhoods covering E where |f(y)-f_n(x_i)(y)|<2e for y in each neighborhood and n>N(x_i). So yes. The N(x_i) depends on x and e. Now take N to be the max{N(x_i)}. It's the max of a finite number of integers.

Last edited: Jan 14, 2009
14. Jan 14, 2009

### tim_lou

Sorry for quoting the wrong post before. And indeed, you missed proving Mn->0. Dick's approach certainly works. However, this above also works, and I think it would help a lot to see this problem from a different angle.

Suppose Mn ≥ ϵ, Mn is always achieved by compactness, say g(xn)=Mn.

consider the set An={x|gn(x)≥ ϵ}

∩An =∅, why?
but gn(xn)=Mn≥ ϵ for all n means ∩An is not empty, why? (consider limit point compactness, say some subsequence xnj->x. Consider gn(x). gn is decreasing is crucial here)

Last edited: Jan 14, 2009
15. Jan 14, 2009

### e(ho0n3

I don't understand how using max{N(x_i)} helps because in the neighborhood B(x_i, N_i, e) where x lives, the only n for which |f_n(x) - f(x)| < 2e is true is N_i.

16. Jan 14, 2009

### Dick

If the convergence is monotone, then if |f_N(x)-f(x)|<2e then |f_n(x)-f(x)|<2e for any n>N. The f_n are increasing toward an upper bound of f. The f_n can only get closer to f as n increases.

17. Jan 14, 2009

### Dick

Ah. I see what you are doing. If Mn>=e for all n, then you've got an intersection of nested nonempty closed sets on a compact space. Sure. That works too.

18. Jan 14, 2009

### e(ho0n3

Got it. Thanks.

19. Jan 14, 2009

### e(ho0n3

No clue.

If ∩An is not empty, there is an x such that gn(x) = Mn for all n. I don't see how this is possible though.

20. Jan 19, 2009

### e(ho0n3

I finally figured this out. If there was an x in the intersection, then gn(x) >= e for all n, but this implies that e <= lim gn(x), which is impossible.

If x is in An, then it is in A(n+1) because g(n+1)(x) >= gn(x) >= e. It follows that A1 contains A2 contains A3 etc. Also, each An is a closed subset by the continuity of gn. By compactness, as Dick noted, there is at least one element contained in each An so the intersection is not-empty.

Thanks a lot tim_lou.